Showing a ball contained in a ball

1. May 21, 2012

trap101

Recall definition of an open ball in Rn centered at a point a with radius r > 0:

"regular definition"

Prove that if r < s then B(r,a)  B(s,a)

Idea: So I've decided to let a pointx be in B(s,a), now I have that r < |X-a| < s

I'm wondering if I should split this up into two steps and show that r < |x - a| and then |x - a| < s. ..........but I'm really stuck on how to do this formally.

2. May 21, 2012

gopher_p

What does it mean for one set to be contained in another? i.e what is the definition of $U\subseteq V$?

3. May 21, 2012

trap101

In terms of this, it means that all the points in U are also contained in V and all the points of V are contained in U. So a B(r,x) is also contained within V and vice versa.

4. May 21, 2012

gopher_p

That's right. And to show that one set $U$ is contained in another set $V$ we usually need to pick an arbitrary element of $U$ and show that it must also be in $V$. Since the sets we work with usually have "membership rules" which define them, this amounts to showing that the membership rules which define $U$ force its members to satisfy the membership rules for $V$.

So here, we're trying to show that $B(r,a)\subseteq B(s,a)$ whenever $r<s$.

What are the membership rules for $B(r,a)$ and $B(s,a)$?

Edit: You edited your post while I was replying. You had the definition right the first time (the one I quoted).

Last edited: May 21, 2012
5. May 21, 2012

Just to chase your idea down for a second, I think what you're trying to do here is to pick a point x in B(s,a) such that r < |x - a| < a right?
Picking that point is not something you prove, you picked it by assumption & you are justified in picking it as you've ensured the point is in the set by requiring
|x - a| < s, right? But ask yourself how picking any radius |x - a| applies to proving that B(r,a) ⊆ B(s,a), how do proofs involving ⊆ use radii? Do they? Can they?

6. May 21, 2012

trap101

Ok so the rules for $B(r,a)$ are that for any point "x" in $B(r,a)$ there exists a ball $B(ρ,a)$ in which the radius of ρ < r. So if I can show that this point is contained within $B(r,a)$ then that would mean that this point is also contained within $B(s,a)$ ?

That actually leads me to another question that has been irking me for awhile, is there any method that could give me a hint of when I'd be using the radii and triangle inequality instead of another method?

7. May 21, 2012

You're probably referring to the proof that an open ball is an open set, where you show that the open ball around any point inside an open ball is contained within the original open ball which is done by exploiting radii & the triangle inequality. In that case what you're trying to do is show that the point is an interior point & the way to do that is by showing that some radius exists such that the ball around the point is contained within the original ball. In this post what you're trying to do is just show that the open ball about a point with radius r is contained within the open ball about the same point of radius a such that r < a. Here the emphasis is on containment, i.e. ⊆, while the other proof's emphasis is on showing the point is an interior point (by constructing a radius small enough so that the ball will fit around the prospective interior point). In other words, you're trying to find the radius in one situation while in this situation you're given the radius. The general method is paying slavish attention to definitions, especially with topological concepts which can be defined in at least 30 different but equivalent ways.

8. May 21, 2012

gopher_p

OK. here is where you are confused. $B(r,a)=\{x:d(x,a)<r\}$ where $d:\mathbf{R}^n\times\mathbf{R}^n\rightarrow \mathbf{R}$ is the distance function. If we are using $|x-a|$ to denote our distance function, then $B(r,a)=\{x:|x-a|<r\}$.

So $|x-a|<r$ is the membership rule for $B(r,a)$.

9. May 21, 2012

trap101

Makes sense. So to show a certain point is an interior, boundary point, etc that's where the use of the radii comes in because I'm choosing the point to represent "all points". Ok I understand what your saying, actually will probably help in my comprehending these questions alot more. Thanks.

10. May 21, 2012

trap101

I think I see what has to be done next:

So I have my point "X" and we have said it's in $B(r,a)$, which also means $|x-a|<r$ now that means that "X" is also in $B(s,a)$ since $|x-a|<r < s ∴$B(r,a) $\subseteq$ B(s,a)##

11. May 21, 2012

gopher_p

This is a "textbook" example of "definition chasing" or "following your nose". Sometimes you just gotta write down precisely what it is that you're given and what you're trying to show, and the rest just falls out.

12. May 21, 2012

trap101

Yeah, seems like I'm trying to make some of these questions harder than they are just because the sometimes "obvious" answer just seems like it may be a little too easy to have figured out. I'm having that problem on the next question I'm working on.

Thanks for the help.